1. 1 Review Parallelogram Properties Review Rectangle Properties Review Square Properties Finding the Area of a Parallelogram Finding the Area of Rectangles PROBLEM 1 PROBLEM 2 PROBLEM 3 PROBLEM 8 PROBLEM 9PROBLEM 10 Standards 7, 8, 10 PROBLEM 4 PROBLEM 5 PROBLEM 7 PROBLEM 6 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. 2 Standard 7: Students prove and use theorems involving the properties of parallel lines cut by a transversal, the properties of quadrilaterals, and the properties of circles. Estándar 7: Los estudiantes prueban y usan teoremas involucrando las propiedades de líneas paralelas cortadas por una transversal, las propiedades de cuadriláteros, y las propiedades de círculos. Standard 8: Students know, derive, and solve problems involving perimeter, circumference, area, volume, lateral area, and surface area of common geometric figures. Estándar 8: Los estudiantes saben, derivan, y resuelven problemas involucrando perímetros, circunferencia, área, volumen, área lateral, y superficie de área de figuras geométricas comunes. Standard 10: Students compute areas of polygons including rectangles, scalene triangles, equilateral triangles, rhombi, parallelograms, and trapezoids. Estándar 10: Los estudiantes computan áreas de polígonos incluyendo rectángulos, triángulos escalenos, triángulos equiláteros, rombos, paralelogramos, y trapezoides. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 3 1. Two pairs of parallel sides. 2. Two pairs of congruent sides. 3. Diagonals bisect each other 4. Opposite angles are congruent. A D B C PARALLELOGRAM 5. Consecutive angles are supplementary. mBmC+=180° mAmB+= mB A B A B B C B C mCmD+= D C D C mDmA+= A D A D Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 1. Two pairs of parallel sides. 2. Two pairs of congruent sides. 5. Diagonals bisect each other 6. Opposite angles are congruent. 7. Consecutive angles are supplementary. mAmB+=180° mBmC+= mCmD+= mDmA+= A D B C 4. Diagonals are congruent 3. All angles are right. RECTANGLE Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 1. Two pairs of parallel sides. 2. All sides are congruent. 5. Diagonals bisect each other 7. Opposite angles are congruent. 8. Consecutive angles are supplementary. mAmB+=180° mBmC+= mCmD+= mDmA+= A D B C 4. Diagonals are congruent 3. All angles are right. 6. Diagonals form a right angle SQUARE Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 6 Area of a Parallelogram B H A = B H where: B= base H= height A= area Could you solve for B? A = B H H H B= A H Could you solve for H? A = B H B B H= A B Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 Find the area of the following parallelogram: 10 5 A = B H where: B= 10 H= 5 A= ? A = 10 5 A = 50 The area is 50 square units Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 What is the height of the following parallelogram if the area is 100 square units? 9 H A = B H where: B= 9 H= ? A= 100 Substituting and solving for H: 100 = 9 H 9 9 H= 100 9 H 11.1 Lineal units Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 9 90 ft 120 ft 72 ft A = B H where: B= base H= height A= area We know that for a parallelogram: H=H= B=B= A= (120 ft)(72ft) A=8640 ft 2 Find the Area: Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 Standards 7, 8, 10 B B-5 H = B-5 A = B H A= 50 H=? B=? 50=B( )B-5 A parallelogram has a height that is 5 in less than its base. If the area is 50 in. Find the measure of the height and the base. 2 50=B -5B 2 -50 0 = B -5B -50 2 (B+5) (B-10) =0 B+5=0 B-10=0 -5 +10 B= -5 B=10 in -5 -50 Two numbers that multiplied be negative fifty should be (+)(-) or (-)(+) Two numbers that added be negative 5 should be |(-)| >| (+)| (1)(-50) 1+(-50)= -49 (5)(-10) 5+(-10)= -5 (2)(-25) 2+(-25)= -23 Now using:First equation:H = B-5 H=( )-510 H=5 in Cheking the solution: A = B H=(10in)(5in)=50in 2 No sense in problem Click for alterante solution to second degree equation 50 = B H First equation Second equation Solving both equations together:

11. 11 0 =1B -5B -50 2 Sometimes the quadratic equation has a prime polynomial expression and can’t be factored or simply, because it is too taugh factoring for you. You may use the Quadratic Formula: X= -b b - 4ac 2a2a 2 +_ where:0 = aX +bX +c 2 Our equation is: a= 1 b= -5 c= -50 We substitute values: B= -( ) ( ) - 4( )( ) 2( ) 2 +_ 1 1-5 -50 B= 5 25 + 200 2 +_ 5 225 B= 2 +_ 5 15 B= 2 + 5 15 B= 2 - 2 20 B= =10 2 -10 B= =-5 5 15 B= 2 +_ + - Both values of B are the same that we got by factoring in the problem. B=10 B= -5 Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 Find the area of the parallelogram if the measure of angle y is: Standards 7, 8, 10 a) 25° b) 35° c) 48° d) 70° y° 20 10 We need to find H to find the area, so we use trigonometry to do it: H Sin y° = H 10 Sin y° = H 10 Solving for H: (10) H= 10sin y° a) y°= 25°.4226 A = B H from the figure: A = 20 10sin y° B = 20 A = 200sin y° A = 200sin ( ) 25° A = 200 ( ) A = 84.52 and So the formula to find the area is: A = 200sin y° Using this formula for each angle: b) y°= 35°.5736 A = 200sin ( ) 35° A = 200 ( ) A = 114.72 c) y°= 48°.7431 A = 200sin ( ) 48° A = 200 ( ) A = 148.62 d) y°= 70°.9397 A = 200sin ( ) 70° A = 200 ( ) A = 187.94 Reflection: What would be the area with a 90°? Which quadrilateral would be this one? How is the area changing in relation to the angle? Trigonometry Ratios Review

13. 13 Find the area of a parallelogram whose vertices are at (-4,-5), (2,-5), (-2,-8), and (4, -8) x y 4 2 6-2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 -10 a) Calculating the base’s length using points (-4,-5) and (2,-5) B = ( - ) + ( - ) 22 = ( 6 ) + ( 0 ) 22 = 36 y 1 y 2 x 1 x 2 2 -4 -5 B= 6 b) The height is | - |= H=3 c) Calculating the area: A = B H where: B= 6 H= 3 A= ? A = ( )( ) 6 3 A= 18 square units Standards 7, 8, 10 d = (x –x ) + (y –y ) 2 2 1 1 2 2 -8-5 3 units 3 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. 14 Area of a Rectangle L A = L W where: W= width L= length A= area Could you solve for W? A = L W L L W= A L Could you solve for L? W A = L W W W L= A W Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15. 15 Find the area of the following rectangle. 7 in A = L W where: W= 4 in L= 7 in A= ? 4 in A = (7in)(4in) A = 28 in 2 The area of the rectangle is 28 square inches. Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16. 16 What is the width of a rectangle if the lenght is 6 cm and the area 9 square centimeters? A = L W where: W= ? L= 6 cm Let’s substitute and solve for W: 6cm W= 9cm 6 W= 1.5 cm 9 cm = ( 6 cm)W 2 Standards 7, 8, 10 W=? 6 cm 9 cm 2 A= 9 cm 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17. 17 The length of a rectangle is 1 cm. less than twice its width and its perimeter is 82 cm. Find the area of the rectangle. L=2W-1 The perimeter of a rectangle is: P = L L W + W W + W L L + P= L +L +W + W P = 2L + 2W 2W-1 W 82=2L+2W first equation Second equation Solving both equations together: L=2W-1 82=2L+2W 82 =2( ) +2W 2W-1 82 = 4W - 2 + 2W 82 = 6W - 2 +2 84 = 6W 6 W= 14 cm L=2( )-1 14 L=28-1 L=27 cm A= (27cm)(14cm) = 378 cm 2 Standards 7, 8, 10 Then the area of the rectangle is: A = L W PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18. 18 4 in 16 in 8 in 4 in A 1 A 2 A 3 A 1 A 2 A 1 16 - 4 - 8=4 in A 1 = (4 in)(4 in) = 16 in 2 A 2 8 in 4 +4 = 8 in A 2 =(8in)(8in) = 4 in 2 2 = 8 in 2 2 = 64 in 2 A 3 4 in 4+4+4=12 in A 3 = (4 in)(12 in) A 3 Total Area= + + = 48 in 2 Total Area= 16 in + 64 in + 48 in 2 2 2 = 128 in 2 What is the Area of a Square? A = s 2 s Find the Area of the figure: Standards 7, 8, 10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19. 19 o u i C B A Tan C= i o Adjacent side Opposite side TANGENT Sin C= i u Hypotenuse Opposite side SINE Cos C= o u Hypotenuse COSINE Adjacent side RETURN TO PROBLEM REVEWING TRIGONOMETRY RATIOS: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved